DIFFERENTIAL EQUATIONS OF THE n"' OrDER. 



If 



then 

 whence 



Thus 



tz Iz 



dx ' chj 



Ex. 1. z ^= Çi{ij -f- .*■) 4(y — •^') (Ç fiii'^l 4 îii"hitraiy) . 



dx^ dxdy dy^ 



p = cj^' 4 _ </, 4' , ,^ = ^- 4 _i_ ^ 4- ^ 



r = </'" 4 — 2'/'' 4' + '/' 4" ' 



« = '/'" 4 — '/'4" > 



f = </)" 4 -f 2cp' 4' + <? 4" • 



f — r = 4 (/)' 4' ) /' + ? = 2(^' 4 , ? — P = 24''/' • 

 Therefore y^ — p^ = 4 (/) 4 </>' 4' and finally q^ — p'' = z (t — r) ; 

 the given primitive is therefore general solution to this difi". equ. 



Ex. 2. z = 2/</)(a-) + xA.iy) + <^{x) 4 (j/) . 



Hence ;j = _?/■/)'+ 4 + 4 "Z*' ' ? = '*■ 4' + '?' + </' 4' ' ^ = </'' + 4' + "/''4' • 

 From the equations r = C'/+4) */'" ' ^ = ('*-'+'/') 4" ""'^ cannot eliminate </>" 

 and 4" ill order to obtain one more equation between x , y , z , (^ , 4 1 



</>', 4'. 



From the expressions for » and (7 we get é' = /JUi 4' = ; 



therefore . = t^ .^ ?Z:^ ^ .(tVOC^t^SP), 



or s Çxy + ?/(/) -j- X -d/ -{- (p 4^) = px -f î?y +/'5" — Oj^ + -"^^^ + ^"4^) 

 or finally s (,v?/ -j- ^) - px -{- gy — z , 

 a partial ditf. equ. of the second order. 



Ex. 3. z = FCcp , -4^) , 



Avhere (/> = </'("), 4 = 4(^') fiutl " and v are given functions of x and 3/, 

 Here we have 





^Z <p' ^J* + ^ 4' ^ 



d(f dy dip dy 



Deriving from these by differentiation r , s and t and eliminating <p" and 



4" , Ave get 



lu tv /in ?v , Tin Tiv\ , _, Sm Sv 



r — — — s -\- — — I -{-r — -J— 



dy dy \dx dy dy dx/ dx dx 



A — (P' 

 d<p ^ 



B^Z 4^' + C -^-'^ CD' 4- 

 dip dydip 



I ...(«) , 



